Proving that Both m and n are Odd using Indirect Proof

[blimkie]

use the method of indirect proof. if m and n are integars and there product mn is odd, prove that both m and n are odd

so i wnat to prove that (m)(n) odd (mn)

so i assume that (mn) is even to go about the indirect method

now i need a jump start i wrote some equations and some algebra but dindt come up with a contradication

help is appreciative that's a lot guys

 

[Pyrrhus]

Start off by

$$m = 2k+1$$

$$n = 2c + 1$$

where k and c is any natural number.

 

[Galileo]

Using contradiction (indirect proof), you can assume either m or n to be even (that is, a multiple of two) and show that mn must also be even.

 

[HallsofIvy]

Start off by
$$m = 2k+1$$
$$n = 2c + 1$$
where k and c is any natural number.

No, that's exactly the wrong thing to do.

Since blimkie wants to use indirect proof to prove that "both m and n must be odd, he should negate that: "either m or n is even".

So assume m= 2p, which is even for any integer p, multiply by n and see what happens!

 

[vaishakh]

A number to be even it should be divisible by 2. thus for the product mn to be divisible by 2 any of the integers m or n must be divisible by prime 2. thus the divisibility testdiectlygives you the answer that then anyone of them must be an even number.